SOLUTION: solve the equation 4^2x+1=2^3x+6 this is what I did. (2^2)^2x+1=2^3x+6 2^4x+1=2^3x+6 4x+1=3x+6 x=5 IS this correct?

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: solve the equation 4^2x+1=2^3x+6 this is what I did. (2^2)^2x+1=2^3x+6 2^4x+1=2^3x+6 4x+1=3x+6 x=5 IS this correct?      Log On


   

Question 62091: solve the equation 4^2x+1=2^3x+6
this is what I did.

(2^2)^2x+1=2^3x+6
2^4x+1=2^3x+6
4x+1=3x+6
x=5
IS this correct?
Found 2 solutions by joyofmath, Earlsdon:
Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
solve the equation 4^2x+1=2^3x+6
You were very close but you calculated that 2*(2x+1) = 4*x+1 when it is in fact 4*x+2.
4%5E%282x%2B1%29+=+2%5E%283x%2B6%29.
2%5E%282%5E%282x%2B1%29%29+=+2%5E%283x%2B6%29.
2%5E%284x%2B2%29+=+2%5E%283x%2B6%29.
4x%2B2+=+3x%2B6.
4x=3x%2B4.
x=4.
Check your answer and you would have caught the mistake.
4%5E%282%284%29%2B1%29+=+4%5E9+=+2%5E18.
2%5E%283%284%29%2B6%29+=+2%5E18.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You almost got it!
4%5E%28%282x%2B1%29%29+=+2%5E%28%283x%2B6%29%29
%282%5E2%29%5E%28%282x%2B1%29%29+=+2%5E%28%283x%2B6%29%29 Recall: 2%282x%2B1%29+=+4x%2B2
2%5E%28%284x%2B2%29%29+=+2%5E%28%283x%2B6%29%29
4x%2B2+=+3x%2B6
x+=+4
Check:
4%5E%28%282%2A4%2B1%29%29+=+4%5E9 = 262144
2%5E%28%283%2A4%2B6%29%29+=+2%5E18 = 262144